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In the adjoining figure, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.

In the adjoining figure, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Mensuration

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Answer

Given,

Radius of the circle = 10 cm

Angle at the centre subtended by a chord AB = 90°.

We know that,

Area of sector OACB = πr2×90360πr^2 \times \dfrac{90}{360}

= 3.14 × 10 × 10 × 14\dfrac{1}{4}

= 3144\dfrac{314}{4}

= 78.5 cm2.

In right angle △OAB,

Area of △OAB = 12\dfrac{1}{2} × OA × OB

= 12\dfrac{1}{2} × 10 × 10

= 50 cm2.

Area of minor segment = Area of sector OACB – Area of △OAB

= 78.5 - 50

= 28.5 cm2.

Area of circle = πr2

= 3.14 × 10 × 10

= 314 cm2.

Area of major segment = Area of circle – Area of minor segment

= 314 - 28.5

= 285.5 cm2.

Hence, area of sector OACB = 78.5 cm2 and area of major segment = 285.5 cm2.

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